Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $z \neq 0$. $n = \dfrac{z - 5}{z^2 - 12z + 35} \div \dfrac{-7z - 21}{z^2 - 7z} $
Explanation: Dividing by an expression is the same as multiplying by its inverse. $n = \dfrac{z - 5}{z^2 - 12z + 35} \times \dfrac{z^2 - 7z}{-7z - 21} $ First factor the quadratic. $n = \dfrac{z - 5}{(z - 7)(z - 5)} \times \dfrac{z^2 - 7z}{-7z - 21} $ Then factor out any other terms. $n = \dfrac{z - 5}{(z - 7)(z - 5)} \times \dfrac{z(z - 7)}{-7(z + 3)} $ Then multiply the two numerators and multiply the two denominators. $n = \dfrac{ (z - 5) \times z(z - 7) } { (z - 7)(z - 5) \times -7(z + 3) } $ $n = \dfrac{ z(z - 5)(z - 7)}{ -7(z - 7)(z - 5)(z + 3)} $ Notice that $(z - 5)$ and $(z - 7)$ appear in both the numerator and denominator so we can cancel them. $n = \dfrac{ z(z - 5)\cancel{(z - 7)}}{ -7\cancel{(z - 7)}(z - 5)(z + 3)} $ We are dividing by $z - 7$ , so $z - 7 \neq 0$ Therefore, $z \neq 7$ $n = \dfrac{ z\cancel{(z - 5)}\cancel{(z - 7)}}{ -7\cancel{(z - 7)}\cancel{(z - 5)}(z + 3)} $ We are dividing by $z - 5$ , so $z - 5 \neq 0$ Therefore, $z \neq 5$ $n = \dfrac{z}{-7(z + 3)} $ $n = \dfrac{-z}{7(z + 3)} ; \space z \neq 7 ; \space z \neq 5 $